Due Mar 2, 2:59 AM EST
We have a dataset of size 8 and we want to separate a subset of size 3 from it. How many ways do we have to do it?
That's right! We have to pick 3 objects out of 8 in the dataset. The order does not matter we cannot pick the same object twice. So we have combinations of size 3 out of 8 objects. There are 8x7x6/(3x2x1)=8x7=56 of them.
We have two disjoint datasets, of size 8 and 5 respectively. We would like to move three objects from the first dataset to the second and simultaneously move two objects from the second dataset to the first one. How many ways do we have to do it?
This is correct. When we pick K objects in the dataset of size N (K=3, N=8 for the first dataset and K=2, N=5 for the second), we are dealing with combinations: unordered selection without repetitions. Thus we can pick 3 objects in the first dataset in 8x7x6/(3x2x1)=8x7=56 ways and we can pick 2 objects from the second dataset in 5x4/(2x1)=10 ways. Since we want to pick a subset from both datasets, we are dealing with pairs of choices, so we have to multiply variants for both datasets.
In a 6 number lottery one is trying to guess an unordered subset of 6 numbers among 44 numbers without repetitions. For this one picks 6 numbers out of 44 himself. How many ways are there to do this? You can use wolfram alpha to compute the exact number.
This is correct! We have to pick a subset of size 6 out of 44 options. These are combinations and the answer is a binomial coefficient.
In a 6 number lottery one is trying to guess an unordered subset of 6 numbers among 44 numbers without repetitions. After the lottery the organisers decided to count how many possible ways are there to guess correctly exactly four numbers. What is the answer to this question? You can use wolfram alpha to compute the exact number.
This is correct! Since the lottery has already happened, the set of winning numbers is fixed. So to guess exactly four winning numbers one has to pick four numbers among 6 winning numbers and pick two other numbers among remaining 44-6=38 numbers. Choices for both cases can be computed as binomial coefficients and then we have to multiply the results by the product rule.
Also note that if we divide this number by the answer to the previous problem we get approximately 0.0015. As we will discuss next week this is the probability to guess exactly four numbers correctly, that is the chances are approximately 0.15%.